When does this inequality have a finite number of integer solutions?

Question

[Choose the one correct answer and explain your reasoning.]

Let $a$ , $b$ , $c$ be positive numbers. There are finitely many positive whole numbers $x$ , $y$ which satisfy the inequality $a^x > cb^y$ if

$a>1$ or $b<1$ .

$a<1$ or $b<1$ .

$a<1$ and $b<1$ .

$a<1$ and $b>1$ .

Suggestion

It’s important to notice that the question refers to a finite number of solutions…

Can we use logarithms to rewrite the inequality?

Solution

Taking logarithms of both sides we have $a^x > cb^y \quad\iff\quad \log(a^x)> \log(cb^y)$ (so long as the base of the logarithm is greater than one, as $\log x$ is then an increasing function).

This can be rewritten as

\begin{align*} x\log a &> \log c+y\log b\\ \iff\quad y\log b &< x\log a - \log c \end{align*}

The corresponding equation is of a straight line in the $x-y$ plane. To find the gradient of this line we want to divide through by $\log b$ but we must be careful as this could be negative.

We have either $\text{Case 1: }\quad y < x\frac{\log a}{\log b} - \frac{\log c}{\log b},\quad\text{if } \log b > 0$ or $\text{Case 2: }\quad y > x\frac{\log a}{\log b} - \frac{\log c}{\log b},\quad\text{if } \log b < 0.$

To have a finite number of positive solutions the inequality must correspond to the region below a straight line with negative gradient.

Case 2 is a region above a line, so we need only consider case 1, $\log b > 0$ . We also require the gradient of the line, $\dfrac{\log a}{\log b}$ , to be negative.

Hence, $\log b > 0$ and $\log a < 0$ . In other words, $b > 1$ and $a < 1$ .

So the answer is (d).

Why does it not matter whether the $y$ -intercept of the straight line is positive or negative?

Three possible straight lines